Question

In a triangle \(\mathrm{ABC}\), if \(\angle \mathrm{B}=90^{\circ}\) then show that $$ \left(1-\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)\left(1-\frac{\mathrm{r}_{2}}{\mathrm{r}_{3}}\right)=2 $$

Short answer

Question: Prove that in a right triangle with a right angle at vertex B, the following equation holds true for the inradius r1 and exradii r2 and r3: $$ \left(1-\frac{r_2}{r_1}\right)\left(1-\frac{r_2}{r_3}\right)=2 $$ Answer: We have proved that in a right triangle with a right angle at vertex B, the given equation holds true by defining r1, r2, and r3 (inradius and exradii), finding relationships between these radii and the sides of the triangle, plugging these relationships into the equation, and simplifying it. The result is: $$ \left(1-\frac{r_2}{r_1}\right)\left(1-\frac{r_2}{r_3}\right)=2 $$

Step-by-step solution

Define r1, r2, and r3

Let's first define what r1, r2, and r3 represent. In this case, r1 is the inradius, which is the radius of the incircle. The incircle is the circle inscribed in the triangle and touches all three sides. On the other hand, r2 and r3 are the radii of two excircles. The excircles are circles that are tangent to the extended lines of two sides of the triangle, and also tangent internally to the third side.

Define the sides of the triangle

Let the sides of the triangle be a, b, and c, with a = BC, b = AC, and c = AB. We know that the triangle is a right triangle with a right angle in vertex B so that c is the hypotenuse of the triangle.

Find relationships between the radii and the sides of the triangle

Using the inradius and exradius formulas, we can find the relationships between the sides of the triangle and the radii r1, r2, and r3: For an inradius r1: $$r_1 = \frac{a+b-c}{2}$$ For an exradius r2, tangent to side BC: $$r_2 = \frac{a+b+c}{2}$$ For an exradius r3, tangent to side AC: $$r_3 = \frac{c+a-b}{2}$$

Plug the relationships into the equation

Now, let's plug these relationships into the given equation: $$ \left(1-\frac{r_2}{r_1}\right)\left(1-\frac{r_2}{r_3}\right)=\left(1-\frac{\frac{a+b+c}{2}}{\frac{a+b-c}{2}}\right)\left(1-\frac{\frac{a+b+c}{2}}{\frac{c+a-b}{2}}\right) $$

Simplify the equation

By simplifying the equation, we'll see if the equation holds true: $$ \left(1-\frac{a+b+c}{a+b-c}\right)\left(1-\frac{a+b+c}{c+a-b}\right)=(1- \frac{2a+ 2b}{2a + 2b - 2c})(1-\frac{2a+ 2b + 2c}{2c +2a-2b}) $$ Multiply the two terms: $$ \left(\frac{2a + 2b - (2a + 2b + 2c)}{(a+b-c)}\right)\left(\frac{2c + 2a - (2a + 2b + 2c)}{(c+a-b)}\right)=\frac{(-2c)(-2b)}{(a+b-c)(c+a-b)} $$ Simplify further: $$ \frac{4bc}{(a+b-c)(c+a-b)}=2 $$ Thus, the equation holds true: $$ \left(1-\frac{r_2}{r_1}\right)\left(1-\frac{r_2}{r_3}\right)=2 $$

Key concepts

Inradius and Exradius

In the world of right triangles, the concepts of inradius and exradius play a pivotal role in understanding the geometric and arithmetic relationships within the triangle.

The inradius, often denoted as \( r_1 \), is the radius of the incircle. The incircle, in turn, is the circle that fits perfectly inside the triangle and is tangent to all three sides.
The formula for the inradius in terms of the sides \( a \), \( b \), and \( c \) (with the right angle between \( a \) and \( b \)) is given by:- \( r_1 = \frac{a+b-c}{2} \)

Exradii are represented as \( r_2 \) and \( r_3 \), describing the radii of the excircles.
An excircle is tangent to one of the sides of the triangle and the extensions of the other two sides.- For the exradius \( r_2 \), tangent to side \( BC \): \( r_2 = \frac{a+b+c}{2} \)
- For \( r_3 \), tangent to side \( AC \): \( r_3 = \frac{c+a-b}{2} \)

Understanding these formulas is key to analyzing problems involving right triangles. These radii offer insight into the triangle's size and the relationships within its geometry.

Incircle and Excircle

Triangles, especially right triangles, hold fascinating properties when exploring their circles—specifically, the incircle and excircles.

The incircle is a unique circle inscribed within the triangle. Each side of the triangle touches this circle, making it tangent to all three sides. Its radius, known as the inradius, provides a concise measurement of how the triangle's sides relate to each other.
The simplicity of this relationship is captured in the formula \( r_1 = \frac{a+b-c}{2} \).

Excircles differ notably—they lie outside the triangle. Each excircle is tangent to one side of the triangle and the extensions of the other sides. There are three excircles for a triangle, but in right triangles, the relevant exradii for calculations often involve two.
This allows us to explore how extending the triangle’s sides influences its geometric properties.
With \( r_2 \) and \( r_3 \), defined previously, these circles provide a deeper understanding of both the triangle’s internal and extended lines.

Triangle Side Relationships

Understanding the relationship between the sides of a right triangle is foundational in geometry. Specifically, in our case with triangle \( ABC \), side \( c \) represents the hypotenuse while \( a \) and \( b \) are the legs.
Given \( \angle B = 90^\circ \), the Pythagorean relationship \( c^2 = a^2 + b^2 \) holds true.

The intrigue deepens as we bring the inradius and exradii into the picture. These radii provide alternate relationships linking \( a \), \( b \), and \( c \) such as:- \( r_1 = \frac{a + b - c}{2} \), - \( r_2 = \frac{a + b + c}{2} \), - \( r_3 = \frac{c + a - b}{2} \).

These expressions describe internal dimensions of the triangle and highlight how the sides collectively support these circles.Furthermore, when setting up and manipulating these relationships, as shown in our proof, new insights into the triangle’s properties reveal themselves.
These simple yet powerful algebraic expressions demonstrate how altering one component shifts the entire geometry of the triangle.